A light wave is traveling linearly in a medium of dielectric constant 4, incidents on the horizontal interface are separating the medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be :

(Given : relative permeability of medium μ_r = 1)

CONCEPT:

• The total internal reflection is defined as the phenomenon that occurs when the light rays travel from a more optically denser medium to a less dense medium.
• The permeability is written as;

           \(μ = √ {μ_rϵ_r}\)

          Here we have μ_r as the relative permeability and ϵ_r as the dielectric constant.

CALCULATION:

Given:

μr = 1, Medium of dielectric constant, ϵr = 4

For the total internal reflection, we know that angle of incidence is greater than the critical angle and it is written as;

i > \(θ_c\)      -----(1)

Now, taking the sine function on both sides in equation (1) we have;

sin i > sinθ_c  

and sinθc = \(\frac{ μ_R}{μ_D}\)

Therefore we have;

sin i > \(\frac{ μ_R}{μ_D}\)     -----(2)

The permeability is equal to  \(μ = √ {μ_rϵ_r}\)

For,  μr = √1×1  

⇒  μr = 1

and,   μD = √4×1  

⇒  μD = 2

Now, from equation (2) we have;

sin i > 1/2

⇒ i > sin^-1 \(\frac{1}{2}\)

⇒ i > 30^o

Now from the above option, i > 30^o is only 60^o.

Hence option 4) is the correct answer.